document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1079691>Question 1079691</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let BAC be a right triangle. O is a point on side AC. D and E are points on side BC. AD is a line and OE is a line. DA intersects BO at point F. ADC is a right triangle. BOE is a right triangle. \r<BR>\n" );
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document.write( "Given AB = OA and OA equals CO, how do I calculate OF/OE?\r<BR>\n" );
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document.write( "This is so hard, my teacher said she doubts anyone can solve it. ;(  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #694029 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=694029\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let k be any positive number, so k > 0. The value k represents the distance from A to B, or the side length AB. Let's make\r<BR>\n" );
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document.write( "A = (0,0)<BR>\n" );
document.write( "B = (0,k)<BR>\n" );
document.write( "C = (2k,0)<BR>\n" );
document.write( "O = (k,0) which is is on segment AC; specifically at the midpoint<BR>\n" );
document.write( "D,E are points on segment BC; unknown locations for now\r<BR>\n" );
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document.write( "Notice how based on those definitions above, we would have<BR>\n" );
document.write( "AB = k<BR>\n" );
document.write( "AC = 2k\r<BR>\n" );
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document.write( "Furthermore, let's make these declarations:<BR>\n" );
document.write( "line p = line through points A and D<BR>\n" );
document.write( "line q = line through points B and O<BR>\n" );
document.write( "line r = line through points O and E<BR>\n" );
document.write( "line s = line through points B and C<BR>\n" );
document.write( "We will use all four lines. They will pair up to help us locate points F and E. We do not need to know the location of D. \r<BR>\n" );
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document.write( "Line p and line q will pair up to find the coordinates of F<BR>\n" );
document.write( "Line r and line s will pair up to find the coordinates of E\r<BR>\n" );
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document.write( "Yes there's a lot to remember here. I recommend drawing everything out to get a good idea of what things look like. See below\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"https://i.imgur.com/FvPgp4M.png\"><BR>\n" );
document.write( "Note: Image generated using <a  rel=nofollow HREF= \"https://www.geogebra.org/\">GeoGebra</a> (free graphing software).\r<BR>\n" );
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document.write( "The red segments are the ones we truly care about, but will take a bit of work to find. \r<BR>\n" );
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document.write( "Take note of the letters next to each line. For instance, a \"p\" is next to the line through points A and D, which denotes that this is line p. Hopefully the labeling isn't too cluttered and you can find each label just fine. \r<BR>\n" );
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document.write( "Let's get started finding the slope of line BC, which will help us find line p eventually.\r<BR>\n" );
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document.write( "slope of line BC:<BR>\n" );
document.write( "m = (y2 - y1)/(x2 - x1)<BR>\n" );
document.write( "m = (0-k)/(2k-0)<BR>\n" );
document.write( "m = (-k)/(2k)<BR>\n" );
document.write( "m = -1/2\r<BR>\n" );
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document.write( "Anything perpendicular to line BC will have a slope of 2/1, or just 2. We find this by flipping the fraction, and flipping the sign, of -1/2. This means that line p has a slope of 2.\r<BR>\n" );
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document.write( "line p has slope m = 2 and it goes through a known point (0,0), so we have (x,y) = (0,0) ---> x = 0 and y = 0\r<BR>\n" );
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document.write( "y = mx+b<BR>\n" );
document.write( "y = 2x+b<BR>\n" );
document.write( "0 = 2*0+b<BR>\n" );
document.write( "b = 0\r<BR>\n" );
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document.write( "Therefore the equation of line p is y = 2x+0 or simply y = 2x\r<BR>\n" );
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document.write( "Now let's find the equation for line q.\r<BR>\n" );
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document.write( "This line is going through two known points B = (0,k) and O = (k,0)\r<BR>\n" );
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document.write( "slope:<BR>\n" );
document.write( "m = (y2 - y1)/(x2 - x1)<BR>\n" );
document.write( "m = (0-k)/(k-0)<BR>\n" );
document.write( "m = (-k)/(k)<BR>\n" );
document.write( "m = -1\r<BR>\n" );
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document.write( "Use one of the points, say B = (0,k), and the slope and we get\r<BR>\n" );
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document.write( "y = mx+b<BR>\n" );
document.write( "k = (-1)*0+b<BR>\n" );
document.write( "k = 0+b<BR>\n" );
document.write( "k = b<BR>\n" );
document.write( "b = k\r<BR>\n" );
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document.write( "So y = mx+b turns into y = -x+k<BR>\n" );
document.write( "If k was a small number, such as k = 1, then<BR>\n" );
document.write( "y = -x+k<BR>\n" );
document.write( "y = -x+1<BR>\n" );
document.write( "which is the equation of line q. \r<BR>\n" );
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document.write( "Now we have the equations <BR>\n" );
document.write( "y = 2x<BR>\n" );
document.write( "y = -x+1<BR>\n" );
document.write( "Use any method you prefer to solve for (x,y) and you'll get (x,y) = (1/3, 2/3). <BR>\n" );
document.write( "line p and line q intersect to form point F, so<BR>\n" );
document.write( "F = (1/3, 2/3)\r<BR>\n" );
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document.write( "Keep this point in mind. The coordinates are very important. \r<BR>\n" );
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document.write( "Note: I skipped the steps on showing how I solved the system above to save time/space. Let me know if you need to see these steps. \r<BR>\n" );
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document.write( "Now we need the equation for line r\r<BR>\n" );
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document.write( "We know the slope of line q is -1 which is the same as -1/1. Flip the fraction and sign and we get 1/1 = 1. The perpendicular slope is 1. The equation of line r will have slope of 1. Why? Because we want lines q and r to be perpendicular, so we can form a 90 degree angle BOE.\r<BR>\n" );
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document.write( "Point O is at (k,0) which is updated to (1,0) if we use k = 1 like above. It's important to stick with the same k value whatever it is you picked earlier.\r<BR>\n" );
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document.write( "Line r goes through (1,0) and the slope is 1, so \r<BR>\n" );
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document.write( "y = mx+b<BR>\n" );
document.write( "y = 1x+b<BR>\n" );
document.write( "0 = 1*1+b<BR>\n" );
document.write( "0 = 1+b<BR>\n" );
document.write( "0-1 = 1+b-1<BR>\n" );
document.write( "b = -1\r<BR>\n" );
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document.write( "The equation for line r is: y = 1x-1 or just y = x-1\r<BR>\n" );
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document.write( "We will use line r later. For now we need to find line s. This line goes through points B and C. Recall that I made k = 1, so \r<BR>\n" );
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document.write( "B = (0,k) = (0,1)<BR>\n" );
document.write( "C = (2k,0) = (2*1,0) = (2,0)\r<BR>\n" );
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document.write( "We need the slope of the line through (0,1) and (2,0)\r<BR>\n" );
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document.write( "slope: <BR>\n" );
document.write( "m = (y2 - y1)/(x2 - x1)<BR>\n" );
document.write( "m = (0 - 1)/(2 - 0)<BR>\n" );
document.write( "m = -1/2\r<BR>\n" );
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document.write( "Slope of line s = slope of segment BC = -1/2\r<BR>\n" );
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document.write( "B = (0,1) is the y intercept of line s, so that means the equation of line s is: y = (-1/2)x + 1\r<BR>\n" );
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document.write( "Line r was found to be y = x-1<BR>\n" );
document.write( "Line s was found to be y = (-1/2)x + 1\r<BR>\n" );
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document.write( "Use any method to solve and you should get (x,y) = (4/3, 1/3)\r<BR>\n" );
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document.write( "Note: I also skipped a few steps similar to the last system. \r<BR>\n" );
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document.write( "At this point, you're probably wondering why I am doing all this work to find these coordinates. The reason I'm doing all this is so we can use the distance formula to find the distance between...<BR>\n" );
document.write( "* Points F and O<BR>\n" );
document.write( "* Points O and E\r<BR>\n" );
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document.write( "That will give the lengths of segments FO and OE respectively. Let's find the distance from F = (1/3, 2/3) to O = (1,0)<BR>\n" );
document.write( "d = sqrt( (x1-x2)^2 + (y1-y2)^2 )<BR>\n" );
document.write( "d = sqrt( (1/3-1)^2 + (2/3-0)^2 )<BR>\n" );
document.write( "d = sqrt( 8/9 )<BR>\n" );
document.write( "d = sqrt(8)/sqrt(9)<BR>\n" );
document.write( "d = sqrt(8)/3<BR>\n" );
document.write( "d = 0.94280904158207<BR>\n" );
document.write( "The last decimal value is approximate<BR>\n" );
document.write( "So FO = 0.94280904158207 approximately\r<BR>\n" );
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document.write( "Similarly, the distance from O = (1,0) to E = (4/3, 1/3) is...<BR>\n" );
document.write( "d = sqrt( (x1-x2)^2 + (y1-y2)^2 )<BR>\n" );
document.write( "d = sqrt( (1-4/3)^2 + (0-1/3)^2 )<BR>\n" );
document.write( "d = sqrt( 2/9 )<BR>\n" );
document.write( "d = sqrt(2)/sqrt(9)<BR>\n" );
document.write( "d = sqrt(2)/3<BR>\n" );
document.write( "d = 0.47140452079103<BR>\n" );
document.write( "The last value shown above is approximate<BR>\n" );
document.write( "So OE = 0.47140452079103 approx\r<BR>\n" );
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document.write( "Now there's one last step: dividing the lengths\r<BR>\n" );
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document.write( "FO/OE = 0.94280904158207/0.47140452079103 = 2.00000000000002<BR>\n" );
document.write( "which is close enough. We have a bit of rounding error with the decimal approximations\r<BR>\n" );
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document.write( "If you ended up using the radical versions, with the square roots, then you'd perfectly end up with a ratio of 2.\r<BR>\n" );
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document.write( "FO/OE = (sqrt(8)/3)/(sqrt(2)/3) = (sqrt(8)/3) divided by (sqrt(2)/3) = 2\r<BR>\n" );
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document.write( "It's pretty interesting how all of this mess leads to a nice clean answer. \r<BR>\n" );
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document.write( "Final Answer: <font color=red size=5>2</font>\r<BR>\n" );
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document.write( " </SPAN>\n" );
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